Derivatives - Local Minima & Maxima

  • Local Minima and Maxima are points on a graph where the function reaches its highest or lowest values within a certain interval.
  • We can find these points by using the derivative of the function.
  • Let’s start with an example to understand how to find local minima and maxima.
  • Example: Find the local minima and maxima of the function f(x) = x^3 - 3x^2.

Step 1: Find the derivative of the function

  • Differentiate f(x) = x^3 - 3x^2 to find its derivative.
  • f’(x) = 3x^2 - 6x.

Step 2: Set the derivative equal to zero to find critical points

  • Set f’(x) = 0 to find the critical points where local minima and maxima may occur.
  • 3x^2 - 6x = 0.
  • Simplifying the equation, we get: x(x - 2) = 0.
  • The critical points are x = 0 and x = 2.

Step 3: Determine the nature of the points

  • To determine the nature of the points, we need to find the second derivative of the function.
  • Let’s find the second derivative of f(x).

Step 4: Find the second derivative

  • Differentiate f’(x) = 3x^2 - 6x to find its second derivative.
  • f’’(x) = 6x - 6.

Step 5: Test the critical points

  • Test each critical point to determine if it is a local minimum, local maximum, or neither.
  • Substitute the critical points into the second derivative equation.
  • For x = 0: f’’(0) = 6(0) - 6 = -6.
  • For x = 2: f’’(2) = 6(2) - 6 = 6.

Conclusion

  • The critical point x = 0 corresponds to a local maximum, and x = 2 corresponds to a local minimum.
  • Therefore, the local minima and maxima of the function f(x) = x^3 - 3x^2 are at x = 2 and x = 0, respectively.

Slide 11:

  • To find the local minima and maxima of a function, we need to analyze the behavior of the derivative.
  • A point where the derivative changes sign from positive to negative represents a local maximum.
  • A point where the derivative changes sign from negative to positive represents a local minimum.
  • A point where the derivative is zero can be a local minimum, local maximum, or neither, depending on the behavior of the second derivative.
  • Let’s look at an example to understand this concept further.

Slide 12:

  • Example: Find the local minima and maxima of the function f(x) = x^2 - 4x + 3.
  • Step 1: Find the derivative of the function.
  • f’(x) = 2x - 4.

Slide 13:

  • Step 2: Set the derivative equal to zero to find critical points.
  • 2x - 4 = 0.
  • Simplifying the equation, we get: x = 2.
  • The critical point is x = 2.

Slide 14:

  • Step 3: Determine the nature of the critical point using the second derivative.
  • Let’s find the second derivative of f(x).
  • Step 4: Find the second derivative.
  • Differentiate f’(x) = 2x - 4 to find its second derivative.
  • f’’(x) = 2.

Slide 15:

  • Step 5: Test the critical point.
  • Substitute the critical point x = 2 into the second derivative equation.
  • f’’(2) = 2.
  • Since the second derivative is positive, the critical point represents a local minimum.

Slide 16:

  • Conclusion for the example: The function f(x) = x^2 - 4x + 3 has a local minimum at x = 2.

Slide 17:

  • We can also use the concept of concavity to determine the nature of the critical points.
  • A function is concave up if its second derivative is positive, and concave down if its second derivative is negative.
  • A local minimum occurs at a point where the function changes from concave down to concave up.
  • A local maximum occurs at a point where the function changes from concave up to concave down.

Slide 18:

  • Let’s look at another example to understand the concavity concept.
  • Example: Find the local minima and maxima of the function f(x) = x^3 - 6x^2 + 9x.

Slide 19:

  • Step 1: Find the derivative of the function.
  • f’(x) = 3x^2 - 12x + 9.

Slide 20:

  • Step 2: Set the derivative equal to zero to find critical points.
  • 3x^2 - 12x + 9 = 0.
  • Simplifying the equation, we get: x = 1.
  • The critical point is x = 1.

Slide 21:

  • Step 3: Determine the concavity of the critical point using the second derivative.
  • Let’s find the second derivative of f(x).
  • Step 4: Find the second derivative.
  • Differentiate f’(x) = 3x^2 - 12x + 9 to find its second derivative.
  • f’’(x) = 6x - 12.

Slide 22:

  • Step 5: Test the critical point.
  • Substitute the critical point x = 1 into the second derivative equation.
  • f’’(1) = 6(1) - 12 = -6.
  • Since the second derivative is negative, the critical point represents a local maximum.

Slide 23:

  • Conclusion for the example: The function f(x) = x^3 - 6x^2 + 9x has a local maximum at x = 1.

Slide 24:

  • We can also have cases where the function has no local minima or maxima.
  • For example, consider the function f(x) = x^2.
  • Let’s analyze this function.

Slide 25:

  • Step 1: Find the derivative of the function.
  • f’(x) = 2x.

Slide 26:

  • Step 2: Set the derivative equal to zero to find critical points.
  • 2x = 0.
  • Simplifying the equation, we get: x = 0.
  • The critical point is x = 0.

Slide 27:

  • Step 3: Determine the concavity of the critical point using the second derivative.
  • Let’s find the second derivative of f(x).
  • Step 4: Find the second derivative.
  • Differentiate f’(x) = 2x to find its second derivative.
  • f’’(x) = 2.

Slide 28:

  • Step 5: Test the critical point.
  • Substitute the critical point x = 0 into the second derivative equation.
  • f’’(0) = 2.
  • Since the second derivative is positive, the critical point represents neither a local minimum nor a maximum.

Slide 29:

  • Conclusion for the example: The function f(x) = x^2 has no local minima or maxima.

Slide 30:

  • In summary, to find local minima and maxima:
    • Find the derivative of the function.
    • Set the derivative equal to zero to find critical points.
    • Determine the concavity of the critical points using the second derivative.
    • Test each critical point to determine if it is a local minimum, local maximum, or neither.
    • Interpret the results accordingly.